I've been working on a proof related to the additivity of Riemann integrals and would greatly appreciate insights and feedback for clarity and correctness of the proof. Because i've never seen a text, that tried to prove in this way. Below is the statement and demonstration of my proof.
Definition Let $ f(x) $ be a continuous function on an interval $ [a, b] $. Then, if the limit $ \lim_{{\| \Delta x_l \| \to 0}} \sum f(c_k) \Delta x_k $ exists, the function $ f(x) $ is Riemann integrable on $ [a, b] $, and it is defined by:
$$ \lim_{{\| \Delta x_l \| \to 0}} \sum f(c_k) \Delta x_k = \int_a^b f(x) \,dx $$
where the definite integral of $ f(x) $ over the interval $ [a, b] $ results in a new function $ g(x) $ calculated on the interval $ [a, b] $, and it can be expressed as:
$$ \int_a^b f(x) \,dx = g(b) - g(a) $$
By means of the definition.
$$ \int_{a}^{b} f(x) \,dx = \lim_{{\| \Delta x_l \| \to 0}} \sum f(c_k) \Delta x_k $$
where $ c_k $ is in $ [a, b] $, and $ \Delta x_k $ is the width of the corresponding subinterval.
Now, we want to show the relationship.
$$ \int_{a}^{c} f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx $$
We can do this by dividing the interval $ [a, c] $ into two subintervals: $ [a, b] $ and $ [b, c] $.
$$ \int_{a}^{c} f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx $$
Now, we apply the definition of the definite integral to each term:
$$ \lim_{{\| \Delta x_l \| \to 0}} \sum f(c_k) \Delta x_k \Bigg|_a^c = \lim_{{\| \Delta x_l \| \to 0}} \sum f(c_k) \Delta x_k \Bigg|_a^b + \lim_{{\| \Delta x_l \| \to 0}} \sum f(c_k) \Delta x_k \Bigg|_b^c $$
Using the additivity of the limit, we can write:
$$ \lim_{{\| \Delta x_l \| \to 0}} \sum f(c_k) \Delta x_k \Bigg|_a^c = \lim_{{\| \Delta x_l \| \to 0}} \sum f(c_k) \Delta x_k \Bigg|_a^b + \lim_{{\| \Delta x_l \| \to 0}} \sum f(c_k) \Delta x_k \Bigg|_b^c $$
And finally, we rewrite this using the notation of definite integrals:
$$ \int_{a}^{c} f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx $$
It seems correct to me; the thesis makes clearly sense if you think of integrals as the area under the curve of a "one dimensional" function (with one variable). You could also prove that by the definition of integral as it follows: $$\int_a^bf(x)dx = F(b) - F(a)$$ where $F(x)$ is the anti derivative of $f(x)$; now by summing and subtracting a quantity equal to zero you get: $$\int_a^bf(x)dx = F(b) - F(c) + F(c) - F(a)$$ but always from definition: $$F(b) - F(c) + F(c) - F(a) = \int_c^bf(x)dx + \int_a^cf(x)dx$$ putting it all together you get the thesis: $$\int_a^bf(x)dx = \int_a^cf(x)dx + \int_c^bf(x)dx$$